UVA 12594 Naming Babies (斜率优化DP)

来源：互联网发布：啥软件机票便宜编辑：程序博客网时间：2024/06/05 00:28

思路：设dp[i][j] 为将前j个字符分成i段的最小值。a[i]为第i个字符在原始字符串的位置。

那么dp[i][j] = min(dp[i][j],dp[i-1][k]+0*a[k+1]+1*a[k+2] + ...... + (j-k+1)*a[j] - a[k+1]^2 - ...... - a[j]^2)

原始等价于k*a[k+1]+(k+1)*a[k+2] + ...... + (j-1)*a[j] - k*(a[k+1] + ....... + a[j]) - a[k+1]^2 - ...... - a[j]^2

用s1[]记录a[i]的和，s2[]记录(i-1)*a[i]的和，s3[]记录a[i]^2的和

然后就可以用斜率优化了。

我的代码：

#include<cstdio>#include<algorithm>#include<iostream>#include<cstring>using namespace std;const int maxm = 505;const int maxn = 20005;int dp[maxm][maxn];int key[30],n,m,a[maxn],s1[maxn],s2[maxn],s3[maxn];int q[maxn],hd,tl;char str[maxn];int getDP(int j,int k,int i){    return dp[i-1][k] + s2[j] - s2[k] - k*(s1[j]-s1[k]) - (s3[j]-s3[k]);}int getY(int j,int k,int i){    return (dp[i-1][k]-s2[k]+k*s1[k]+s3[k]) - (dp[i-1][j]-s2[j]+j*s1[j]+s3[j]);}int getX(int j,int k){    return k - j;}int solve(){    for(int j=1;j<=n;j++) dp[1][j] = s2[j] - s3[j];    for(int i=2;i<=m;i++){        hd = tl = 0;        q[tl++] = i-1;        for(int j=i;j<=n;j++){            while(hd+1<tl && getY(q[hd],q[hd+1],i) <= s1[j]*getX(q[hd],q[hd+1])) hd++;            dp[i][j] = getDP(j,q[hd],i);            while(hd+1<tl && getY(q[tl-1],j,i)*getX(q[tl-2],q[tl-1])                  <= getY(q[tl-2],q[tl-1],i)*getX(q[tl-1],j)) tl--;            q[tl++] = j;        }    }    return dp[m][n];}int main(){    int cas;    scanf("%d",&cas);    for(int T = 1; T <= cas; T++){        scanf("%s%d",str,&m);        for(int i=0;i<strlen(str);i++){            key[str[i]-'a'] = i;        }        scanf("%s",str);n = strlen(str);        a[1] = key[str[0]-'a'];        s1[1] = a[1];s2[1] = 0;s3[1] = a[1]*a[1];        for(int i=1;i<n;i++){            a[i+1] = key[str[i]-'a'];            s1[i+1] = s1[i] + a[i+1];            s2[i+1] = s2[i] + i*a[i+1];            s3[i+1] = s3[i] + a[i+1]*a[i+1];        }        printf("Case %d: %d\n",T,solve());    }    return 0;}

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